Question

If $\left(x+\frac{1}{x}\right)=4$, find the values of
(i) $\left(x^2+\frac{1}{x^2}\right)$ and
(ii) $\left(x^4+\frac{1}{x^4}\right)$.

Answer 1

$$\begin{gathered} \left(i\right)\left(x+\frac{1}{x}\right)=4 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2={\left(4\right)}^2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)\right)=16 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)+2=16 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=16-2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=14 \end{gathered}$$

Therefore, the value of x²+$\frac{1}{x^2}$ is 14.

$$\begin{gathered} \left(x^2+\frac{1}{x^2}\right)=14 \\ Squaringboththesides: \\ \Rightarrow \left(x^4+\frac{1}{x^4}+2\left(x^2\right)\left(\frac{1}{x^2}\right)\right)={\left(14\right)}^2 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)+2=196 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)=196-2 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)=194 \end{gathered}$$

Therefore, the value of x⁴ + $\frac{1}{x^4}$ is 194.